Grid homology is independent of the grid and computes knot Floer homology

Use the standard genus-one splitting of the oriented three-sphere: the horizontal circles bound parallel meridian disks in one solid torus, and the vertical circles bound parallel meridian disks in the other; their primitive meridians meet once. Taking all n parallel disks cuts each side into n balls. The row and column marking condition puts one O and one X in every corresponding cut ball. Push the horizontal O-to-X segments into the lower side and the vertical X-to-O segments into the upper side, using their evident rectangular bridge disks. These simultaneous arcs give exactly the oriented knot of the grid, since every vertical segment passes over every horizontal segment. Thus this is an actual balanced 2n-pointed diagram of K, not only an abstract torus count. Its analytic symmetric rank is $1+n-1=n$, and w=O,z=X. Give it the actual flat integrable data of [F1]. The identity on states identifies its full polynomial count matrix with the algorithmic one.

Check the remaining good-endpoint condition before applying [F3]. Positivity gives a nonnegative domain D for every actual integrable strip. Its index $n_x(D)+n_y(D)$ is nonnegative. If it is zero, every cell touching either state has coefficient zero. The local equation $\mathrm{\Delta }D=1_x-1_y$ then forces x=y, since at any changed vertex the four coefficients would have nonzero alternating sum. Consequently $\mathrm{\Delta }D=0$, and telescoping its row and column differences gives $D_{ij}=a_i+b_j$. Because x is one point in every row and column, the n cells immediately northeast of its state points have total coefficient ${\sum }_i{a}_i+{\sum }_j{b}_j=0$. The sum of ALL cell coefficients is n times that sum, hence zero. Nonnegativity gives D=0. The actual ordered-root energy bound is then zero, so the strip is constant. Therefore no nonconstant index-at-most-zero strip exists. Finiteness, index-one regularity and square zero were already supplied by [F1,F5]; no index-two regularity of this special integrable path is assumed. The precise endpoint hypotheses of [F3] hold.

Apply the actual full-minus auxiliary and diagram equivalences of [F3] to this endpoint. They identify its chosen-action restriction with a doubly pointed full complex, and identify its ordinary collapse with that complex tensor $V^{\otimes (n-1)}$. They also give the quotient-by-all-variables and associated-graded identities by specializing the actual finite maps and homotopies. At this stage use ONLY the relative formulas of [F1]: on the connected state set, its numerical J-gradings differ from the transported normalized analytic origins by one integer Maslov constant b and one integer Alexander constant c. In particular, if H denotes normalized doubly pointed associated-graded zero-U homology, the grid tilde homology has the relative identification $H\otimes V^{\otimes (n-1)}$ with common shifts (b,c). This constructs the comparison before determining either shift; no assertion of the theorem under proof is a premise.

Fix b without using Alexander centering. List the O markings in order of increasing row, with distinct column indices $o_0,\dots ,o_{n-1}$. Replace X by $X^{\ast }$ having in row i the column $o_{i+1}$, with cyclic indices. This is disjoint from O and is a one-component grid: its marking permutation sends row i to row i+1. Its vertical segments join successive row heights, except for the one segment from the last row to the first. The short vertical segments meet no other horizontal height, so every crossing lies on that sole long vertical segment, which passes OVER the complementary embedded planar arc. This is an unknot: raise the long segment into an upper ball and lower the complementary planar arc into a lower ball; both are boundary-parallel arcs, with the rectangular upper bridge disk and the planar lower bridge disk. Match their boundary chords on the separating plane and join the two disks along that chord to obtain an embedded spanning disk, rounding its finite corners. The FULL unfiltered rectangle matrix and M_O do not depend on X, so replacing X by $X^{\ast }$ changes neither this matrix nor its Maslov grading. By the purely combinatorial graded chosen-action equivalences and move generation in [F2], this same unfiltered complex is graded equivalent to a size-two unknot grid. In that grid the two states have M_O=0,-1; direct rectangle enumeration gives the cone ${\mathbb{F}}_2[U_1,U_2]\stackrel{U_1+U_2}{\to }{\mathbb{F}}_2[U_1,U_2]$ with source degree -1 and target degree zero. The monic polynomial retraction in [F3] evaluates the other variable at the chosen one, so after quotienting that chosen U its total homology is one-dimensional in M_O degree zero. The combinatorial inverse homotopies specialize too, so this is true for the original grid's chosen-action total zero-U complex. Step 3.1 compares it to the analytic total hat homology, whose normalized degree is zero by [F4]. Its common Maslov shift is therefore b=0. This also proves nontriviality without treating the Alexander-associated-graded complex as total hat homology.

Fix c by a finite reciprocal symmetry. Define the finite Laurent polynomial $P_G(t)={\sum }_x(-1{)}^{M_O(x)}{t}^{A_G(x)}$ and put $D_n(t)=(1-t^{-1}{)}^{n-1}$. The Euler characteristic of a finite bigraded complex equals that of its homology degree by degree: split each finite degree into boundaries, a homology complement and a complement mapped isomorphically to boundaries; the latter pairs cancel with opposite Maslov signs. Thus [F2] implies that $Q_G=P_G\mathrm{/}{D}_n$ is unchanged by every grid move, as a rational function: commutation preserves P and stabilization multiplies P by $1-t^{-1}$, precisely the Euler polynomial of V. Construct $G^T$ by transposing the square and INTERCHANGING O and X. The ambient rotation $(x,y,z)\mapsto (y,x,-z)$ has determinant plus one and changes old vertical overpasses to new horizontal underpasses. An old horizontal O-to-X arc becomes a new vertical X-to-O arc; an old vertical X-to-O arc becomes a new horizontal O-to-X arc. Hence it carries the SAME oriented knot to $G^T$, rather than a mirror or assumed orientation reversal. Both grids have size n, so [F2] gives $P_{G^T}=P_G$. Transposition preserves the strict pair relation defining J, while mark exchange swaps M_O and M_X. The literal formulas are therefore $A_{G^T}(x^T)=-A_G(x)-(n-1)$ and $M_{O^T}(x^T)=M_O(x)-2A_G(x)-(n-1)$. Since A_G is integral, their signs give $P_G(t)=(-1{)}^{n-1}{t}^{-(n-1)}{P}_G(t^{-1})$. Using $D_n(t^{-1})=(-1{)}^{n-1}{t}^{n-1}{D}_n(t)$ yields $Q_G(t)=Q_G(t^{-1})$. This derives reciprocity from actual J formulas and purely combinatorial moves; no analytic grading identification or unproved Alexander determinant formula has been used.

Write $Q_H(t)={\sum }_{m,a}(-1{)}^m\dim H_{m,a}{t}^a$. This is a finite Laurent polynomial. By [F4]'s ACTUAL conjugation, $H_{m,a}$ is isomorphic to $H_{m-2a,-a}$. The degree difference is even, hence $Q_H(t)=Q_H(t^{-1})$. For clarity, this symmetry is supplied by the normalized analytic coordinate-swap map: on the vertical corner $(i,j)=(0,a)$ it lands on the horizontal corner $(a,0)$, and multiplying by $U^a$ identifies that corner with $(0,-a)$ and lowers Maslov degree by 2a. It is a map of the corner subquotients, not a declaration that the full positive-z hat differential preserves A. Moreover $Q_H(1)=1$. Indeed the associated graded and total zero-U doubly pointed complexes have the same finite chain groups with their Maslov degrees, so their ordinary Euler characteristics coincide; the latter has exactly one homology class of normalized degree zero. Therefore Q_H is NONZERO. The relative comparison in step 3.1 and b=0 now give $P_G=t^c{D}_n{Q}_H$, or $Q_G=t^c{Q}_H$. Substituting the two reciprocal identities yields $(t^c-t^{-c})Q_H=0$ in the Laurent polynomial domain $\mathbb{Z}[t,t^{-1}]$. Since Q_H is nonzero, c=0. This is precisely the missing absolute combinatorial Alexander anchor, with no appeal to grid equality to establish its own shift.

The identity in step 1.1 and the actual analytic equivalences in step 3.1 consequently preserve both normalized origins, by steps 4.1 and 5.1. Compose with [F3]'s normalized comparison to the diagram H in the Statement; its inverse and inverse homotopies remain finite and filtered. Taking the Alexander associated graded of these identities gives the claimed associated-graded minus equivalences. Each term retains every individual O weight until the explicitly stated restriction or quotient. The ordinary collapse and all-variable-zero associated-graded specialization are exactly the two actual constructions of [F3], yielding respectively $CF{K}^{-}(K)\otimes V^{\otimes (n-1)}$ and $\widehat{HFK}(K)\otimes V^{\otimes (n-1)}$, with the displayed bidegrees now absolutely centered. The chosen-action restriction yields $CF{K}^{-}(K)$ instead. Composing the comparisons for two grids gives their normalized filtered equivalence and inverse homotopies. Finally [F4] identifies the preserved Alexander origin with the zero-framed capped-Seifert half-Chern expression of the Statement. Its cap and negative-Morse convention are inherited from that supplier, not replaced by a winding-number convention. Every original conclusion follows with all positive z terms and all stated scalar distinctions retained.